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                <h1 id="welcome-to">Welcome to 知远数学分享<a class="headerlink" href="#welcome-to" title="Permanent link">&para;</a></h1>
<p><span class="twemoji"><svg xmlns="http://www.w3.org/2000/svg" viewBox="0 0 24 24"><path d="M22 2s-7.64-.37-13.66 7.88C3.72 16.21 2 22 2 22l1.94-1c1.44-2.5 2.19-3.53 3.6-5 2.53.74 5.17.65 7.46-2-2-.56-3.6-.43-5.96-.19C11.69 12 13.5 11.6 16 12l1-2c-1.8-.34-3-.37-4.78.04C14.19 8.65 15.56 7.87 18 8l1.21-1.93c-1.56-.11-2.5.06-4.29.5 1.61-1.46 3.08-2.12 5.22-2.25 0 0 1.05-1.89 1.86-2.32z"/></svg></span> 分享优秀题目与解题方法</p>
<h2 id="_1">题目分享<a class="headerlink" href="#_1" title="Permanent link">&para;</a></h2>
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<li><strong>典型题目</strong> - 通过题目可以得出一个一般性结论，更快更好的解题</li>
<li><strong>方法总结</strong> - 求解一类题目的常用方法</li>
<li><strong>知识总结</strong> - 汇总常用公式（不等式、三角函数、函数图形与性质）</li>
<li><strong>经典的错误，标准的零分</strong> - 总结常见的易错点并分析原理</li>
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<h2 id="_2">今日更新<a class="headerlink" href="#_2" title="Permanent link">&para;</a></h2>
<p><code>2021-7-18</code> </p>
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<p><strong>题目来源：</strong> 武钟祥 - 《高等数学辅导讲义》P53 例2</p>
<p><strong>题目：</strong> 设 <span class="arithmatex">\(f(0) = 0\)</span>，则 <span class="arithmatex">\(f(x)\)</span> 在点 <span class="arithmatex">\(x=0\)</span> 可导的 <strong>充要条件</strong> 为：</p>
<p>（A）<span class="arithmatex">\(\lim\limits_{h \to 0} {\cfrac{1}{h^2}f(1-\cos h) }\)</span> 存在</p>
<p>（B）<span class="arithmatex">\(\lim\limits_{h \to 0} {\cfrac{1}{h}f(1-e^h) }\)</span> 存在</p>
<p>（C）<span class="arithmatex">\(\lim\limits_{h \to 0} {\cfrac{1}{h^2}f(h-\sin h) }\)</span> 存在</p>
<p>（D）<span class="arithmatex">\(\lim\limits_{h \to 0} {\cfrac{1}{h}  \left [ f(2h) - f(h) \right ]}\)</span> 存在</p>
<details class="note"><summary>答案与解析</summary><p>答案：B</p>
<div class="tabbed-set" data-tabs="1:4"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><label for="__tabbed_1_1">A选项</label><div class="tabbed-content">
<div class="arithmatex">\[\begin{align}\lim_{h \to 0}\cfrac{f(1-\cos h)}{h^2} &amp;= \lim_{h \to 0}\cfrac{f(1-\cos h) - f(0)}{1-\cos h}\cdot \cfrac{1-\cos h}{h^2}\\&amp;= \cfrac{1}{2}\lim_{h \to 0}\cfrac{f(1-\cos h) - f(0)} {1-\cos h}=\cfrac{1}{2}f'_+(0).\end{align}\]</div>
<p>由于当 <span class="arithmatex">\(h \to 0\)</span> 时，<span class="arithmatex">\((1-\cos h) \to 0^+\)</span>，则只能推出右导数存在，故A不正确。</p>
</div>
<input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><label for="__tabbed_1_2">B选项</label><div class="tabbed-content">
<div class="arithmatex">\[\begin{align}\lim_{h \to 0} \cfrac{1}{h}f(1-e^h)  &amp;= \lim _{h \to 0}\cfrac{f(1-e^h)-f(0)}{1-e^h}\cdot \cfrac{1-e^h}{h}\\&amp;= -\lim _{h \to 0} \cfrac{f(1-e^h)-f(0)}{1-e^h} = -f'(0).\end{align}\]</div>
</div>
<input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><label for="__tabbed_1_3">C选项</label><div class="tabbed-content">
<div class="arithmatex">\[\begin{align}\lim_{h \to 0} \cfrac{1}{h^2}f(h-\sin h) &amp;=\lim_{h \to 0} \cfrac{f(h-\sin h) - f(0)}{h - \sin h}\cdot \cfrac{h-\sin h}{h^2}.\end{align}\]</div>
<p>由于 <span class="arithmatex">\(\lim\limits_{h \to 0} \dfrac{h-\sin h}{h^2} = 0\)</span>，极限C存在时， <span class="arithmatex">\(\lim\limits_{h \to 0}\dfrac{f(h-\sin h) - f(0)}{h - \sin h}\)</span> 不一定存在，故C选项错误</p>
</div>
<input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><label for="__tabbed_1_4">D选项</label><div class="tabbed-content">
<p>D选择错误类型与其他选项不同，<span class="arithmatex">\(f(2h)-f(h)\)</span> 是两个动点，不符合导数定义。反例如下：</p>
<p>取<span class="arithmatex">\(f(x) = \left\{\begin{matrix} 1 ,x \ne 0\\  0 , x = 0 \end{matrix}\right.\)</span>，显然 <span class="arithmatex">\(f'(0)\)</span> 不存在，因为 <span class="arithmatex">\(f(x)\)</span> 在 <span class="arithmatex">\(x = 0\)</span> 处不连续，但该极限存在。</p>
</div>
</div>
</details>
<details class="danger"><summary>总结</summary><p><span class="arithmatex">\(f'(x_0)\)</span> 存在 <span class="arithmatex">\(\to  \lim \dfrac{f(x_0 + \Box )-f(x_0)}{\Box}\)</span> 存在</p>
<ol>
<li><span class="arithmatex">\(\Box \to 0\)</span></li>
<li><span class="arithmatex">\(\Box \ne 0\)</span></li>
</ol>
<p><span class="arithmatex">\(\lim \dfrac{f(x_0 + \Box )-f(x_0)}{\Box}\)</span> 存在 <span class="arithmatex">\(\to f'(x_0)\)</span> 存在</p>
<ol>
<li><span class="arithmatex">\(\Box \to 0\)</span></li>
<li><span class="arithmatex">\(\Box \ne 0\)</span></li>
<li><span class="arithmatex">\(\Box\)</span> 可正可负</li>
</ol>
<p><span class="arithmatex">\(\lim\limits_{h \to 0} \dfrac{f(\varphi(h))-f(0)}{\psi(h)}\)</span> 存在 <span class="arithmatex">\(\to f'(0)\)</span> 存在</p>
<ol>
<li><span class="arithmatex">\(\varphi(x) \to 0，\varphi(x) \ne 0\)</span></li>
<li><span class="arithmatex">\(\varphi(x)\)</span> 可正可负</li>
<li><span class="arithmatex">\(\varphi(x)\)</span> 与 <span class="arithmatex">\(\psi(x)\)</span> 同阶</li>
</ol>
</details>
<hr />
<h2 id="markdown">题目MarkDown模板<a class="headerlink" href="#markdown" title="Permanent link">&para;</a></h2>
<div class="highlight"><pre><span></span><code><span class="gs">**题目来源：**</span> 武钟祥 - 《高等数学辅导讲义》P53 例2

<span class="gs">**题目：**</span> 设 $f(0) = 0$，则 $f(x)$ 在点 $x=0$ 可导的 <span class="gs">**充要条件**</span> 为：

（A）$\lim\limits_{h \to 0} {\cfrac{1}{h^2}f(1-\cos h) }$ 存在

（B）$\lim\limits_{h \to 0} {\cfrac{1}{h}f(1-e^h) }$ 存在

（C）$\lim\limits_{h \to 0} {\cfrac{1}{h^2}f(h-\sin h) }$ 存在

（D）$\lim\limits_{h \to 0} {\cfrac{1}{h}  \left [ f(2h) - f(h) \right ]}$ 存在

??? note &quot;答案与解析&quot;
    答案：B

    === &quot;A选项&quot;

        $$\begin{align}\lim_{h \to 0}\cfrac{f(1-\cos h)}{h^2} &amp;= \lim_{h \to 0}\cfrac{f(1-\cos h) - f(0)}{1-\cos h}\cdot \cfrac{1-\cos h}{h^2}\\&amp;= \cfrac{1}{2}\lim_{h \to 0}\cfrac{f(1-\cos h) - f(0)} {1-\cos h}=\cfrac{1}{2}f&#39;_+(0).\end{align}$$

        由于当 $h \to 0$ 时，$(1-\cos h) \to 0^+$，则只能推出右导数存在，故A不正确。

    === &quot;B选项&quot;
        $$\begin{align}\lim_{h \to 0} \cfrac{1}{h}f(1-e^h)  &amp;= \lim <span class="ge">_{h \to 0}\cfrac{f(1-e^h)-f(0)}{1-e^h}\cdot \cfrac{1-e^h}{h}\\&amp;= -\lim _</span>{h \to 0} \cfrac{f(1-e^h)-f(0)}{1-e^h} = -f&#39;(0).\end{align}$$

    === &quot;C选项&quot;
        $$\begin{align}\lim_{h \to 0} \cfrac{1}{h^2}f(h-\sin h) &amp;=\lim_{h \to 0} \cfrac{f(h-\sin h) - f(0)}{h - \sin h}\cdot \cfrac{h-\sin h}{h^2}.\end{align}$$

        由于 $\lim\limits_{h \to 0} \dfrac{h-\sin h}{h^2} = 0$，极限C存在时， $\lim\limits_{h \to 0}\dfrac{f(h-\sin h) - f(0)}{h - \sin h}$ 不一定存在，故C选项错误

    === &quot;D选项&quot;
        D选择错误类型与其他选项不同，$f(2h)-f(h)$ 是两个动点，不符合导数定义。反例如下：

        取$f(x) = \left\{\begin{matrix} 1 ,x \ne 0\\  0 , x = 0 \end{matrix}\right.$，显然 $f&#39;(0)$ 不存在，因为 $f(x)$ 在 $x = 0$ 处不连续，但该极限存在。

??? danger &quot;总结&quot;

    $f&#39;(x_0)$ 存在 $\to  \lim \dfrac{f(x_0 + \Box )-f(x_0)}{\Box}$ 存在

    <span class="k">1.</span> $\Box \to 0$
    <span class="k">2.</span> $\Box \ne 0$

    $\lim \dfrac{f(x_0 + \Box )-f(x_0)}{\Box}$ 存在 $\to f&#39;(x_0)$ 存在

    <span class="k">1.</span> $\Box \to 0$
    <span class="k">2.</span> $\Box \ne 0$
    <span class="k">3.</span> $\Box$ 可正可负

    $\lim\limits_{h \to 0} \dfrac{f(\varphi(h))-f(0)}{\psi(h)}$ 存在 $\to f&#39;(0)$ 存在

    <span class="k">1.</span> $\varphi(x) \to 0，\varphi(x) \ne 0$
    <span class="k">2.</span> $\varphi(x)$ 可正可负
    <span class="k">3.</span> $\varphi(x)$ 与 $\psi(x)$ 同阶

---
</code></pre></div>
                
              
              
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